The distance of the point (1, 1, 9) from the point of intersection of the line ${{x - 3} \over 1} = {{y - 4} \over 2} = {{z - 5} \over 2}$ and the plane x + y + z = 17 is :

Given, P(1, 1, 9).<br/><br/>Equation of plane x + y + z = 17<br/><br/>Equation of line $\Rightarrow$ ${{x - 3} \over 1} = {{y - 4} \over 2} = {{z - 5} \over 2}$<br/><br/>$\Rightarrow$ ${{x - 3} \over 1} = {{y - 4} \over 2} = {{z - 5} \over 2} = \lambda$ (let)<br/><br/>$\Rightarrow$ x = $\lambda$ + 3; y = 2$\lambda$ + 4; z = 2$\lambda$ + 5<br/><br/>$\therefore$ The point we have is ($\lambda$ + 3, 2$\lambda$ + 4, 2$\lambda$ + 5).<br/><br/>$\because$ This point lies on the plane x + y + z = 17.<br/><br/>$\therefore$ $\lambda$ + 3 + 2$\lambda$ + 4 + 2$\lambda$ + 5 = 17<br/><br/>$\Rightarrow$ $\lambda$ = 1<br/><br/>$\therefore$ The coordinate of point is (4, 6, 7)<br/><br/>$\therefore$ Required distance between (1, 1, 9) and (4, 6, 7) is<br/><br/>$= \sqrt {{{(4 - 1)}^2} + {{(6 - 1)}^2} + {{(7 - 9)}^2}}$<br/><br/>$= \sqrt {9 + 25 + 4} = \sqrt {38}$